Problem 20 - Essential Calculus

Question

Find the critical numbers of \(f\) (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. $$ f(x)=-\left(x^{2}+8 x+12\right) $$

Step-by-Step Solution

Answer

The function has a single critical number at \(x = -4\). It is increasing on the interval \((-∞, -4)\) and decreasing on the interval \((-4, ∞)\). Therefore, \(x = -4\) is a relative maximum.

Step 1: Find the Derivative of \(f(x)\)

Calculate the derivative of the function: \(f'(x)\) = -2x - 8

Step 2: Find the Critical Numbers

Set the first derivative equal to zero and solve for \(x\). So \(0 = -2x - 8\), thus \(x = -4\) are the critical numbers.

Step 3: Test Intervals

Pick test points in the intervals split by the critical numbers to determine whether the function is increasing or decreasing. Here we can pick -5 for the interval \((-∞, -4)\) and -3 for the interval \((-4, ∞)\). Plugging -5 into the derivative yields a positive result, and plugging -3 yields a negative result. Hence the function is increasing on the interval \((-∞, -4)\) and decreasing on the interval \((-4, ∞)\).

Step 4: Identify Relative Extrema

Due to the function increasing before \(x = -4\) and decreasing after, \(x = -4\) represents a relative maximum.

Step 5: Graph Verification

Graph the original function to confirm these findings. The function should increase up to \(x = -4\) and decrease thereafter, and have a peak at \(x = -4\).

Key Concepts

Derivative

The concept of a derivative is central to understanding how a function behaves. It represents the rate at which a function is changing at any given point. For the function \( f(x) = -(x^2 + 8x + 12) \), finding the derivative, denoted \( f'(x) \), involves applying basic differentiation rules. We use the power rule and the constant rule: the derivative of \( x^2 \) is \( 2x \), the derivative of \( x \) is 1, and constants like 12 become 0.Thus, when differentiating \( f(x) \) and applying the negative sign from \( -(x^2 + 8x + 12) \), we obtain:

• \( f'(x) = -2x - 8 \)

The derivative \( f'(x) \) provides a formula to calculate the slope at any point \( x \). This helps us determine critical numbers and analyze the behavior of the function over intervals.

Relative Extrema

Relative extrema refer to the points on a graph where the function reaches a local maximum or minimum. These points are crucial for understanding the shape and peaks of a graph. To find these points, we first identify the critical numbers. This is done by setting the derivative \( f'(x) = -2x - 8 \) equal to zero and solving for \( x \). This gives us \( x = -4 \), marking it as a critical number.To determine if \( x = -4 \) is a relative maximum or minimum, we examine how the function behaves around this point:

• If the function changes from increasing to decreasing at a critical number, it signifies a relative maximum.
• Conversely, if it changes from decreasing to increasing, it's a relative minimum.

In our case, since the function increases as \( x \) approaches -4 from the left and decreases after \( x \) passes -4, \( x = -4 \) is a point of relative maximum.

Increasing and Decreasing Intervals

Identifying intervals where a function is increasing or decreasing is key to understanding its overall behavior. These intervals offer insights into the function's trends. To determine this, we use test points derived from critical numbers to analyze the sign of the derivative in specific intervals.For the function \( f(x) \), we found the critical number \( x = -4 \). The derivative \( f'(x) = -2x - 8 \) helps us test intervals to the left and right of \( x = -4 \):

• In the interval \((-\infty, -4)\), such as by testing \( x = -5 \), the derivative yields a positive result, indicating the function is increasing.
• Conversely, in the interval \((-4, \infty)\), testing with \( x = -3 \) produces a negative result, suggesting that the function is decreasing.

By understanding these intervals, we confirm that \( f(x) \) increases up to \( x = -4 \) and decreases afterwards, aligning with our findings of \( x = -4 \) being a relative maximum.